We want to find real numbers \(\alpha\) and \(\beta\) satisfying the following: \begin{equation*}\begin{bmatrix} 1 \\ 0\end{bmatrix} = \alpha \begin{bmatrix} 1 \\ 1\end{bmatrix} + \beta \begin{bmatrix} -1 \\ 1 \end{bmatrix}.\end{equation*} In other words, we want to solve the system \begin{align*} 1 & = \alpha - \beta \\ 0 & = \alpha + \beta\end{align*} for \(\alpha\) and \(\beta\).

Adding the equations gives \(2\alpha = 1\), implying that \(\alpha = \frac{1}{2}.\) It follows that \(\beta = -\frac{1}{2}\). Hence, \begin{equation*}\begin{bmatrix} 1 \\ 0\end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 \\ 1\end{bmatrix} + \left(-\frac{1}{2}\right)\begin{bmatrix} -1 \\ 1 \end{bmatrix}.\end{equation*}

It suffices to show that for every choice of \(z_1, z_2\in \mathbb{C}^2\), there exist \(\alpha,\beta\in\mathbb{C}\) such that \(\alpha \begin{bmatrix} i \\ 0 \end{bmatrix} + \beta \begin{bmatrix} 1 \\ 1-i \end{bmatrix} = \begin{bmatrix} z_1 \\ z_2\end{bmatrix}\).

The tuple equation can be written as the system \( \begin{bmatrix} i & 1 \\ 0 & 1-i\end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \begin{bmatrix} z_1 \\ z_2\end{bmatrix}.\)

The augmented matrix for the system is \( \left [ \begin{array}{cc|c} i & 1 & z_1 \\ 0 & 1-i & z_2 \end{array} \right] \).

One can proceed to perform row reduction. But row 2 represents the equation \((1-i)\beta = z_2\). Hence, \(\beta = \frac{z_2}{1-i} = \frac{1+i}{2}z_2\).

Putting this into the equation corresponding to row 1 gives us that \(i\alpha + \frac{1+i}{2}z_2 = z_1\), implying that \(\alpha = -i\left(z_1 - \frac{1+i}{2}z_2\right) = -iz_1 + \frac{-1+i}{2}z_2\). Hence, there is a solution no matter what \(z_1, z_2\) are.

(There is actually no need to solve for \(\alpha\) and \(\beta\) explicitly. The coefficient matrix is square and has determinant \(i(1-i)\neq 0\). By Cramer's rule, a unique solution exists and that is all we need to make our conclusion.)

We want to determine if there exist \(a,b,c \in\mathbb{R}\) such that \[ \begin{bmatrix} 1& 0 \\ 0 & 1\end{bmatrix} = a\begin{bmatrix} 3 & 1 \\ 1 & -1 \end{bmatrix} +b\begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix} +c\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \] We can rewrite the right-hand to obtain \[ \begin{bmatrix} 1& 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix} 3a + c & a + b + c \\ a + b + c & -a +2b \end{bmatrix} \]

Comparing the entries on both sides, we get the system \begin{eqnarray*} 3a + c & = & 1 \\ a + b + c & = & 0 \\ -a + 2b & = & 1 \end{eqnarray*} Note that the top-right and bottom-left entries give the same equation \(a+b+c=0\).

The third equation gives \(a = 2b-1\). Substituing this into the first equation gives \(6b+c = 4\). Adding the last two equations gives \(3b +c = 1\). Subtracting this from \(6b+c=4\) gives \(3b = 3\), implying that \(b = 1\). Hence, \(a = 1\) and \(c = -2\). So there is a solution and the answer to the question is “yes”.

Clearly, the span is simply the set of all linear combinations of the three matrices. But we need to give as simple a description as possible. So we should look at what kind of \(2\times 2\) matrices we are getting.

The span is given by \begin{eqnarray*} && \left \{ \alpha A + \beta B + \gamma C : \alpha,\beta, \gamma \in \mathbb{R} \right \}\ & = & \left \{ \begin{bmatrix} \alpha - \beta & \alpha + \gamma \\ \alpha + \beta + \gamma & 0 \end{bmatrix} : \alpha,\beta, \gamma \in \mathbb{R} \right \}\ \end{eqnarray*} We claim that the set is the same as \(\left \{ \begin{bmatrix} r & s \\ t & 0 \end{bmatrix} : r, s, t \in \mathbb{R} \right \}\)

Hence, we want to show that the system \[\begin{array}{rl} \alpha - \beta = r & (1)\\ \alpha + \gamma = s & (2)\\ \alpha + \beta + \gamma = t & (3) \end{array}\] has a solution for any choice of \(r,s,t\in \mathbb{R}\).

Now \((1)+(3)\) gives \(2\alpha + \gamma = r + t\). Subtracting \((2)\) from this gives \(\alpha = r - s + t\). Substituting this into \((1)\) gives \(\beta = -s + t\) and substituing into \((2)\) gives \(\gamma = -r + 2s -t\). Hence, there is a solution and so the span is given by matrices of the form \(\begin{bmatrix} r & s \\ t & 0 \end{bmatrix}\) where \(r, s, t \in \mathbb{R}\).

It suffices to exhibit a vector in \(P_2\) that cannot be written as a linear combination of \(x\) and \(x^2 - 1\).

We claim that \(x^2\) is such a vector. Suppose that there exist real numbers \(\alpha\) and \(\beta\) such that \(x^2 = \alpha x + \beta (x^2 - 1)\). Hence, \[x^2 = \beta x^2 + \alpha x - \beta.\] Comparing the coefficients of \(x^2\) on both sides gives \(\beta = 1\). Now, the left-hand side has a \(0\) constant term but the constant term on the right-hand side is \(-\beta\), implying that \(\beta = 0\). Hence, \(\beta\) must be \(1\) and \(0\) at the same time, which is impossible.

By definition, \(x^2 + 1\) is in the span of \(\{x^2 + x + 1, x+2\}\) if and only if there exist real numbers \(\alpha\) and \(\beta\) such that \[x^2 + 1 = \alpha (x^2 + x + 1) + \beta (x+2).\] Rewriting the right-hand side, we obtain \[x^2 + 1 = \alpha x^2 + (\alpha+ \beta) x + (\alpha + 2\beta).\] Now, comparing the coefficients of \(x^2\), \(x\), and the constant term on both sides, we get that \begin{eqnarray*} 1 & = & \alpha \\ 0 & = & \alpha + \beta \\ 1 & = & \alpha + 2\beta. \end{eqnarray*} So \(\alpha = 1\). The second equation then gives \(\beta = -1\). But that means \(\alpha + 2\beta = -1 \neq 1\). Hence, \(\alpha\) and \(\beta\) cannot exist and so \(x^2 + 1\) is not in the span of \(\{x^2 + x + 1, x+2\}\).

Clearly, \(W \subseteq \mathbb{R}^3\). To show that \(W\) is a proper subspace of \(\mathbb{R}^3\), we find a vector in \(\mathbb{R}^3\) that is not in \(W\). That is, we want to find \(\alpha,\beta,\gamma\in \mathbb{R}\) so that \(\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix}\) cannot be written as a linear combination of \(\begin{bmatrix} 1 \\ -1 \\ 2\end{bmatrix}\), \(\begin{bmatrix} 2 \\ 1 \\ 1\end{bmatrix}\), \(\begin{bmatrix} 1 \\ 2 \\ -1\end{bmatrix}\), and \(\begin{bmatrix} 0 \\ 1 \\ -1\end{bmatrix}\),

In other words, we seek \(\alpha,\beta,\gamma\in \mathbb{R}\) so that the following has no solution: \begin{align*} \lambda_1 \begin{bmatrix} 1 \\ -1 \\ 2\end{bmatrix} + \lambda_2 \begin{bmatrix} 2 \\ 1 \\ 1\end{bmatrix} + \lambda_3 \begin{bmatrix} 1 \\ 2 \\ -1\end{bmatrix} + \lambda_3 \begin{bmatrix} 0 \\ 1 \\ -1\end{bmatrix} = \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix}. \end{align*} The system can be rewritten as \begin{align*} \begin{bmatrix} 1 & 2 & 1 & 0 \\ -1 & 1 & 2 & 1\\ 2 & 1 & -1 & -1 \end{bmatrix} \begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \lambda_4 \end{bmatrix} = \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix}. \end{align*}

The augmented matrix for the system is therefore \[\left[\begin{array}{cccc|c} 1 & 2 & 1 & 0 & \alpha \\ -1 & 1 & 2 & 1 & \beta \\ 2 & 1 & -1 & -1 & \gamma \end{array}\right]\] Adding row one to row two and subtracting two times row one from row three gives \[\left[\begin{array}{cccc|c} 1 & 2 & 1 & 0 & \alpha \\ 0 & 3 & 3 & 1 & \alpha + \beta \\ 0 & -3 & -3 & -1 & -2 \alpha + \gamma \end{array}\right]\] Adding row to to row three gives \[\left[\begin{array}{cccc|c} 1 & 2 & 1 & 0 & \alpha \\ 0 & 3 & 3 & 1 & \alpha + \beta \\ 0 & 0 & 0 & 0 & -\alpha + \beta + \gamma \end{array}\right]\] If we choose \(\alpha = \beta = 0\) and \(\gamma = 1\), the system will have no solution. Thus, we have shown that \(\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}\) is not in \(W\) and so \(W\) is a proper subspace of \(\mathbb{R}^3\).

Remark. In fact, every vector \(\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix}\) such that \(-\alpha + \beta + \gamma \neq 0\) will not be in \(W\).

We show that \(W\) is not a subspace of \(W_2\). For instance, the constant polynomial \(1 \in W\) since \(1 = 0x^2 + 0x + 1\) and the coefficients add up to 1 which is greater than 0. However, multiplying it by the scalar \(-1\) gives \(-1\), which is not in \(W\).

Remark. Interestingly, \(W\) is closed under vector addition. Pick arbitrary \(u, v \in W\). Then \(u = a_1 x^2 + b_1 x + c_1\) and \(v = a_2 x^2 + b_2 x + c_2\) for some real numbers \(a_1,a_2,b_1,b_2,c_1,c_2\) such that \(a_1+b_1+c_1\geq 0\) and \(a_2+b_2+c_2 \geq 0\). Note that \begin{align*} u + v & = a_1x^2 + b_1x + c_1 + a_2x^2 + b_2x + c_2 \\ & = (a_1 + a_2)x^2 + (b_1 + b_2)x + (c_1 + c_2) \\ \end{align*} Setting \(a = a_1+a_2\), \(b = b_1+b_2\), \(c = c_1 + c_2\) gives \(u+v = ax^2 + bx + c\). Now, \(a+b+c= a_1+a_2+ b_1+b_2+c_1 + c_2 = (a_1+b_1+c_1)+(a_2+b_2+c_2 \geq 0.\) Hence, \(u+v \in W\).